[wp-hackers] Sharing Users/authentication
chris.lott at gmail.com
Thu Aug 30 21:44:21 GMT 2007
So, I cleared everything out (de-activated plugin, deleted rows from
wp_options, copied fresh plugin files) and tried again and here is
what is happening... slightly different now.
Starting with two blogs without the plugin. I install the plugin on
the first (master) blog. Activate it. No errors. When I look at the
options it is showing the right database and prefix. Note: I can't
logout of that blog, though... when I try to sign out I just get
redirected to the same page in the control panel area I am on.
At this point I see only one entry in the wp_options table.
Go to second (slave) blog. Activate plugin. Go to options, put in
database name (it is a diff database) and prefix (which is the same).
No errors. I can sign out.
Back to master blog, still no errors, can;t sign out.
Back to slave blog, can't sign in with any usernames from the Master
blog (says invalid username), old sign-ins from the slave blog still
In the master blog the wp_options line has this value:
The slave blog has this:
No errors showing, but nothing seems to be happening. It's mystifying.
On 8/30/07, Stephen Rider <wp-hackers at striderweb.com> wrote:
> Hi --
> First off, This isn't my plugin, it's Curt Woodard's. I've been
> helping with testing and making suggestions, though, so I'm pretty
> familiar with it. :)
> Are the two blogs on the same or different database?
> Same or different username/ password?
> Somehow it seems that the writing of options isn't working
> correctly. This may be a database problem of some sort. I don't see
> any problems with the related code (yet).
> When exactly do you get these errors? immediately when you activate
> the plugin? When you change the settings?
> On Aug 30, 2007, at 2:51 PM, Chris Lott wrote:
> > Stephen:
> > I'm trying this on a faculty blog and it worked fine on the first
> > blog... but on the second it gives this when I try to activate the
> > plugin:
> > WordPress database error: [Incorrect table name 'usermeta']
> > SELECT * FROM ``.usermeta WHERE user_id = '1' AND meta_key LIKE 'wp_%'
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